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emmasim [6.3K]
3 years ago
15

Simple physics (Final) (Pic provided)

Physics
1 answer:
Igoryamba3 years ago
3 0

Answer:

15m/s

Explanation:

Divide distance by time

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Element 2 explanation just trust
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10.0 ppm 10.4 ppm 10.2 ppm 10.8 ppm 10.1 ppm 10.5 ppm 10.1 ppm Using the new instrument on the first day the technicians got a v
iren [92.7K]

Answer:

The datapoint 9.0 ppm is outlier at the 90% confidence level.

Explanation:

The old data has following values

mean=10.5 mm

standard deviation 0.2 mm

Now the mean of new values is calculated as following

mean=\frac{10+10.4+10.2+10.8+10.1+10.5+10.1}{7}\\mean=10.3 ppm

So the value as 9.0 ppm can be considered easily as outlier in  this regard.

3 0
3 years ago
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

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What is the magnitude of the momentum change of two gallons of water (inertia about 7.3 kg ) as it comes to a stop in a bathtub
aliya0001 [1]

We know that the change in momentum is equals to the product of force and time that is impulse (  F \times t). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,

s=ut+\frac{1}{2} gt^2

Here, u is initial velocity which is zero.

s= \frac{1}{2} gt^2 \\\\ t = \sqrt{\frac{2s}{g} }.

Thus, impulse

= F \times \sqrt{\frac{2s}{g} }

From Newton`s second law,

F =mg

Therefore, impulse

= mg \times \sqrt{\frac{2s}{g} } = m \sqrt{2gs}

Given,  m = 7.3 kg and s = 2.0 m

Substituting these values, we get

Change in momentum = impulse  

= 7.3 \ kg \sqrt{2 \times 9.8 \ m/s^2 \times 2.0 \ m } = 45 .7 \ Ns.

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How are the spiral arms of the milky way detected?
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The spiral structure emerges when galactic clusters (open), H II regions and O & B type stars (young stars) are used as tracers. We know this to be true as other pinwheel galaxies exhibit the same patterns across these tracers as in the milky way.
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3 years ago
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